I wrote a program to:
Print the new array of elements
Print the sum of all elements (or integers)
Actually, I got it right, however, the little problem is, I want to maintain all the duplicates (still within the range of four largest elements). Here's what I mean:
Take an array of numbers: [4,5,-2,3,1,2,6,6]
The four largest numbers are 4,5,6,6. And their sum is 4+5+6+6=21
What the code is doing (not good):
Instead of getting "6,6,5,4" as (described above), the code is printing "6,5,4,3" with the sum as 18.
ALSO, when there are only four elements [with or without duplicates] as in [1,1,1,-5],
let it just add ALL elements. You guessed it, the sum of all elements is -2
How do I order the program to print the necessary duplicate(s) to make the four largest integers?
Here's my code...
var arr = new Array(4,5,-2,3,1,2,6,6);
// var arr = new Array(1,1,1,-5);
// var largArr = new Array();
function largest() {
largArr = Array(0, 0, 0, 0)
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[0]) {
largArr[0] = arr[i];
}
}
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[1] && arr[i] < largArr[0]) {
largArr[1] = arr[i];
}
}
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[2] && arr[i] < largArr[1]) {
largArr[2] = arr[i];
}
}
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[3] && arr[i] < largArr[2]) {
largArr[3] = arr[i];
}
}
console.log(largArr[0], largArr[1], largArr[2], largArr[3]);
console.log(largArr[0] + largArr[1] + largArr[2] + largArr[3]);
}
largest();
I believe there is a genius out there who can help me solve this :)
You could get the top four and filter the original array.
function get4Largest(array) {
const top4 = [...array].sort((a, b) => b - a).slice(0, 4);
return array.filter(v => {
if (top4.includes(v)) {
top4.splice(top4.indexOf(v), 1);
return true;
}
});
}
console.log(get4Largest([4, 5, -2, 3, 1, 2, 6, 6]));
A different approach by keeping indices.
function get4Largest(array) {
return array
.map((v, i) => [v, i])
.sort(([a], [b]) => b - a)
.slice(0, 4)
.sort(([, a], [, b]) => a - b)
.map(([v]) => v);
}
console.log(get4Largest([4, 5, -2, 3, 1, 2, 6, 6]));
If you want sum of largest four numbers then you can easily do with sort, slice and reduce:
numbers.sort((a, b) => b - a).slice(0, 4).reduce((acc, curr) => acc + curr, 0)
const numbers = [4, 5, -2, 3, 1, 2, 6, 6];
const result = numbers
.sort((a, b) => b - a)
.slice(0, 4)
.reduce((acc, curr) => acc + curr, 0);
console.log(result);
You can use a reduce, sort and slice methods of an array like so:
function sumMaxValues (maxLength, values) {
return values
.sort((v1, v2) => v1 > v2 ? -1 : 1)
.slice(0, maxLength)
.reduce((sum, v) => sum + v, 0)
}
console.log(
sumMaxValues(4, [4, 5, -2, 3, 1, 2, 6, 6, 10]),
)
Edit: I fixed, a bug that #gog pointed out. The root cause of a problem was that sort when invoked without a compareFn then "the array elements are converted to strings, then sorted according to each character's Unicode code point value."(sort docs)
If for some reason you want to have a classical type of solution, that avoids modern javascript methods, here's one
const arr = Array(4, 5, -2, 3, 1, 2, 6, 6);
//const arr = Array(1, 1, 1, -5);
function largest(){
const largArr = Array(-1/0, -1/0, -1/0, -1/0);
for(let i = 0; i < arr.length; i++){
for(let j = 0; j < largArr.length; j++){
if(arr[i] > largArr[j]){
for(let k = largArr.length - 2; k >= j; k--){
largArr[k + 1] = largArr[k];
}
largArr[j] = arr[i];
break;
}
}
}
let sum = 0;
for(let j = 0; j < largArr.length; j++){
if(largArr[j] === -1/0){
largArr[j] = 0;
}
sum += largArr[j];
}
console.log(largArr, sum);
}
largest();
-1/0 stands for minus infinity (there can be no smaller number); you may also use Number.NEGATIVE_INFINITY for it. If it's too exotic for your needs, replace -1/0 with any number you are certain is less than any possible number in the array (that however, cannot be zero, since you allow negative numbers also).
I am trying to Write a function that a given array consisting of N integers, return the minimum among all integers which are multiples of 11.
let arr = [-6, -91, 1011, -100, 84, -22, 0, 1, 473]
function solution(A) {
let mult = A.filter(x => x % 11 == 0 && x / 11 >=0 && x / 11 < 1000).map(x => x / 11)
return mult
}
solution(arr)
function solution(arr){
let value = arr[i]
for (let i=0; i< arr.length;i++){
if(arr[i]%11 == 0)
{
if(arr[i] < value)
{
value = arr[i]
}
}
}
return value
}
The question should be broken into two parts, then solved.
First, get all the numbers that are divisible by 11, then find the minimum of them.
So, something like:
function getSmallestDivisibleBy11(arr) {
let divisibleBy11 = []
for (let i = 0; i < arr.length; i++){
if (arr[i] % 11 === 0){
divisibleBy11.push(arr[i])
}
}
let returnVal = Infinity;
for (let i = 0; i < divisibleBy11.length; i++){
if (divisibleBy11[i] < returnVal){
returnVal = divisibleBy11[i]
}
}
return returnVal
}
Logic
From the list of input array, filter the array where the number is divisible by 11. Condition number % 11 === 0.
This will give an array of numbers which are divisible by 11. Here it will be [-22, 0, 473].
To make pick the minimum value. Just sort the array and select the zeroth node of sorted array. Since the array of numbers divisible by 11 are already sorted the value for sorted array will be [-22, 0, 473] and the least value of them will be -22
let arr = [-6, -91, 1011, -100, 84, -22, 0, 1, 473]
function solution(list) {
const divisibleArray = list.filter(x => x % 11 == 0); // This will return all numbers which are divisible by 11
const sortedDivisibleArray = divisibleArray.sort((a, b) => a - b); // Sort the array. The minimum value will be zero th node of sorted array
return sortedDivisibleArray.length > 0 ? sortedDivisibleArray[0] : null;
}
const result = solution(arr);
console.log(result);
Single Line Solution
let arr = [-6, -91, 1011, -100, 84, -22, 0, 1, 473];
const solution = (list) => list.filter(x => x % 11 == 0).sort((a, b) => a - b)[0] || null;
const result = solution(arr);
console.log(result);
you can try this
function solution(A) {
var minElement = 100001;
arr.forEach(element => {
let e = element % 11;
if (e === 0 && e < minElement) {
minElement = element;
}
});
console.log(minElement);
return minElement;
}
You can first filter out multiples of 11 from the array and find the minimum among them.
You can use a for...of loop as given below
function solution(A) {
let mult11s = A.filter(x => x % 11 == 0);
let minMult11 = Number.POSITIVE_INFINITY;
for(const mult11 of mult11s){
minMult11 = Math.min(minMult11, mult11);
}
return minMult11;
}
or you can use reduce if you are comfortable with that.
function solution(A) {
return A
.filter(x => x % 11 == 0)
.reduce((acc, curr) => Math.min(acc, curr), Number.POSITIVE_INFINITY);
}
Introduction
The problem is explained on Link to challenge for best instructions
As far as I understand, the idea is to swap elements on the left side of an array with elements from the right side of an array if the element on the left side is greater than 0.
I.e. [2, -4, 6, -6] => [-6, -4, 6, 2].
Positives on the left side have to get swapped with their corresponding "out of place" opposite on the right side. So, unfortunately we can't use
array = array.sort((a, b) => a - b);
In the end, there should be only negatives numbers on the left side and positives on the right side. If the array has an odd length, then one "side" will be one (or more) elements larger, depending on whether or not there are more positives or negatives in the original array.
I.e. [8, 10, -6, -7, 9, 5] => [-7, -6, 10, 8, 9, 5]
Approach
To solve this, I created a pretty lengthy algorithm that splits the array into halfs, and keeps track of "out of place" (positive) elements on the left side and "out of place" (negative) elements on the right side, to use for swapping:
function wheatFromChaff (values) {
//IF ORIGINAL VALUES LENGTH IS EVEN
if (values.length % 2 === 0) {
let left = values.slice(0, values.length / 2);
let right = values.slice(values.length / 2);
let outOfPlaceLeft = left.filter((element) => element > 0);
let outOfPlaceRight = right.filter((element) => element < 0);
//replace positive "out of place" left elements with negative "out of place" right elements
for (let i = 0; i < left.length; i++) {
if (left[i] > 0) {
left.splice(i, 1, outOfPlaceRight.pop());
}
}
//push remaining "out of place" negative right elements to the left
while (outOfPlaceRight.length) {
let first = outOfPlaceRight.shift();
left.push(first);
}
//filter out any negatives on the right
right = right.filter((element) => element > 0).concat(outOfPlaceLeft);
//concat the remaining positives and return
return left.concat(right);
}
//IF ORIGINAL VALUES LENGTH IS ODD
if (values.length % 2 !== 0) {
let left2 = values.slice(0, Math.floor(values.length / 2));
let right2 = values.slice(Math.ceil(values.length / 2));
let middle = values[Math.floor(values.length/2)];
let outOfPlaceLeft2 = left2.filter((element) => element > 0);
let outOfPlaceRight2 = right2.filter((element) => element < 0);
//replace "out of place", positive left elements
for (let j = 0; j < left2.length; j++) {
//if out of there are out of place elements on the right
if (outOfPlaceRight2.length) {
if (left2[j] > 0) {
left2.splice(j, 1, outOfPlaceRight2.pop());
}
}
//if out of place elements on the right are empty
if (!outOfPlaceRight2.length && middle < 0) {
if (left2[j] > 0) {
left2.splice(j, 1, middle);
}
}
}
//filter out negatives on the right
right2 = right2.filter((element) => element > 0);
//unshift remaining "out of place" positive left elements to the right
while (outOfPlaceLeft2.length) {
let first = outOfPlaceLeft2.shift();
right2.unshift(first);
}
if (middle > 0) {
right2.unshift(middle);
}
if (middle < 0) {
left2.push(middle);
}
return left2.concat(right2);
}
}
console.log(wheatFromChaff([2, -6, -4, 1, -8, -2]));
Sometimes the previous approach works, but it gives me an error on Codewars:
AssertionError [ERR_ASSERTION]: [ -10, 7 ] deepEqual [ -10, -3, 7 ]
Do you see any flaws with my logic? Any suggestions for a better strategy?
I have solved this using two "pointers" one starting on the head of the array, and the other starting at the tail of the array. The idea is, on every iteration, to increment the head or decrement the tail iteratively until you found a positive number on the head pointer and a negative number at the tail pointer. When this condition is satisfied, then you have to do a swap on the positions of the elements and continue. Finally, you have to stop the loop when the head pointer is greater than the tail pointer.
function wheatFromChaff(values)
{
let res = [];
let head = 0, tail = values.length - 1;
while (head <= tail)
{
if (values[head] < 0)
{
res[head] = values[head++];
}
else if (values[tail] > 0)
{
res[tail] = values[tail--];
}
else
{
res[tail] = values[head];
res[head++] = values[tail--];
}
}
return res;
}
console.log(wheatFromChaff([2, -4, 6, -6]));
console.log(wheatFromChaff([8, 10, -6, -7, 9]));
.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}
This approach passed all test mentioned on that link:
Time: 894ms Passed: 5 Failed: 0
But maybe there is a better solution.
Another way, using indexOf() and lastIndexOf() helper functions:
function lastIndexOf(arr, pred, start) {
start = start || arr.length - 1;
for (let i = start; i >= 0; i--) {
if (pred(arr[i])) return i;
}
return -1;
}
function indexOf(arr, pred, start = 0) {
for (let length = arr.length, i = start; i < length; i++) {
if (pred(arr[i])) return i;
}
return Infinity;
}
function wheatFromChaff(values) {
let firstPos = indexOf(values, x => x > 0);
let lastNeg = lastIndexOf(values, x => x < 0);
while ( firstPos < lastNeg ) {
let a = values[firstPos];
let b = values[lastNeg];
values[firstPos] = b;
values[lastNeg] = a;
firstPos = indexOf(values, x => x > 0, firstPos);
lastNeg = lastIndexOf(values, x => x < 0, lastNeg);
}
return values;
}
console.log(wheatFromChaff([2, -6, -4, 1, -8, -2]));
I've been struggling for a little while to find or figure out algorithm.
The task:
Basically, I have an array of probabilities:
var input = [0.1, 0.2, 0.3, 0.1];
Let's name these inputs accordingly to: A, B, C and D.
And I also have a variable "m", which can tell me how many of these things needs to happen in order to get the result.
For example:
var m = 2;
This variable m is telling me that the event will happen if any of those two (or more) probabilities will happen.
So in this case, for event to happen, all possible ways for event to happen is:
ABCD
ABC
ABD
BCD
AB
AC
AD
BC
BD and CD
Now I need to calculate their probabilities, which I already have algorithms to calculate AND and OR (where input is just a probabilities array).
AND:
if (input.length > 0) {
output = 1;
}
for (i = 0; i < input.length; i++) {
output = input[i] * output;
}
OR:
if (input.length > 0) {
output = input[0];
}
for (i = 1; i < input.length; i++) {
output = (output + input[i]) - (output * input[i]);
}
So I am struggling on figuring out how to loop through all possible possibilities... And to have something like:
(A and B and C and D) or (A and B and C) or (A and B and D)... and so on... I hope you get the idea.
Here's a simple non-recursive solution to enumerate all combinations with at least m elements.
range = n => [...Array.from({length: n}).keys()]
mask = xs => b => xs.filter((_, n) => b & (1 << n))
at_least = n => xs => xs.length >= n
//
a = [...'ABCD']
m = 2
result = range(1 << a.length).map(mask(a)).filter(at_least(m))
console.log(result.map(x => x.join('')))
Since JS bit arithmetics is limited to 32 bits, this only works for m < 32.
You could get the combinations of your desired array with a minimum of two by using a recursive function which generates all the possible combinations.
function getC(array, min) {
function iter(i, temp) {
var t = temp.concat(array[i]);
if (i === array.length) return;
iter(i + 1, t);
iter(i + 1, temp);
if (t.length >= min) {
result.push(t);
}
}
var result = [];
iter(0, []);
return result;
}
var input = [0.1, 0.2, 0.3, 0.1];
console.log(getC(input, 2).map(a => a.join(' ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
You could go over all combimations of 2 elements (AB, CD etc.) with two nested loops:
for(let i = 0; i < input.length; i++) {
for(let j = i + 1; j < input.length; j++) {
// possible combination: i and j
for(let k = j; k < input.length; k++) {
// possible combination: i, j, k
// and so on
}
}
}
for at least m elements that can be generalized with a nested generator that generates an array of indices ([0, 1], [0, 2], [1, 2], [0, 1, 2]):
function* combinations(length, m = 1, start = 0) {
// Base Case: If there is only one index left, yield that:
if(start === length - 1) {
yield [length - 1];
return;
}
// Otherwise go over all left indices
for(let i = start; i < length; i++) {
// And get all further combinations, for 0 that will be [1, 2], [1] and [2]
for(const nested of combinations(length, m - 1, i + 1)) {
// Yield the nested path, e.g. [0, 1], [0, 1, 2] and [0, 2]
yield [i, ...nested];
}
// If the minimum length is already reached yield the index itself
if(m <= 1) yield [i];
}
}
Now for every combination, we just have to multiply the probabilities and add them up:
let result = 0;
for(const combination of combimations(input.length, m))
result += combination.reduce((prev, i) => prev * input[i], 1);
I am attempting to write a "binary search" which I've never done before. The code below does not work when the value searched for is 6 or 2 and I want to know what I am doing wrong and how to remedy it.
EDIT
To explain what it is suppose to do (based on my understanding) a binary search requires that an array is already sorted, it then looks for the mid-point index of an array. For example, if an array had nine indexes (0-8)the the mid point would be index 4.
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
The algorithm then determines if that mid point has a higher or lower value than the number you are searching for. All elements on the side of the array that does not contain the searched for number and that exist before the midpoint value simply get removed. If the search for value is 8 then the result would be:
[ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
array midpoint value: 5
[ 5, 6, 7, 8, 9 ]
array midpoint value: 7
[ 7, 8, 9 ]
array midpoint value: 8
Code
//_________________________________________________BEGIN notes
// Step 1. Get length of array
// Step 2. Find mid point
// Step 3. Compare if mid point is lower or higher than searched number
// Step 4. lop off unneeded side
// Step 5. go to step 1
//_________________________________________________END notes
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 44, 55];
function getMidPoint(arr, searchNumb) {
var length = arr.length;
var midPoint = Math.floor(length / 2);
var newArr = arr;
console.log(arr);
console.log("array midpoint value: " + arr[midPoint]);
if (arr[midPoint] > searchNumb) {
var newArr = arr.slice(0, arr[midPoint]);
return getMidPoint(newArr, searchNumb);
} else if (arr[midPoint] < searchNumb) {
var newArr = arr.slice(midPoint, arr.length);
return getMidPoint(newArr, searchNumb);
} else {
return arr
}
}
Language agnostic, here is the simplified flow of a recursive binary search implementation, assuming we have an (initially non-empty) array [ARR] and a target [T], where we refer to the middle element of ARR as M:
// 1. If M == T, return true
// 2. If length of ARR is 0, return false (note: step 1 short circuits, ensuring we only hit step 2 if step 1 evaluates to false)
// 3. If T < M, return the result of the recursion on the lower half of ARR
// 4. If T > M, return the result of the recursion on the the latter half of ARR
Following is solution that executes the control flow outlined above. This is similar to solutions already presented in this post, with a few noteworthy differences:
function binarySearch(arr, target, start=0, stop=(arr.length-1)) {
let midPoint = Math.floor(((stop-start)/2) + start)
switch (true) {
case arr[midPoint] === target:
return true
case stop - start === 0:
return false
case arr[midPoint] < target:
return binarySearch(arr, target, midPoint+1, stop)
case arr[midPoint] > target:
return binarySearch(arr, target, start, midPoint)
}
}
Let's unpack the main differences of this implementation:
Slice is no longer used:
We are eschewing the use of Array.prototype.slice because it is a relatively expensive operation (copying half of the current array with each recursive call!) and it is not required for the algorithm to function properly.
In place of slice, we are passing the start and stop indexes of the range of the array that we have narrowed the search down to. This keeps our heap happy by not cluttering it with (potentially many) partial, impermanent copies of the same (potentially massive) array.
We are passing two additional arguments, and they have defaults:
These arguments (start and stop) serve to keep track of the range of the array we are currently recurring on. They are our alternative to slice!
The default arguments enable us to call this recursive function exactly the same as we would when using slice (should the user not provide an explicit range when it is first called).
We are using a switch statement:
The speed of a switch statement vs. an if-else chain depends on several factors, most notably the programming language and the amount of conditionals in each. A switch statement was used here primarily for readability. It is a control flow that matches what we are concerned with handling in this recursive function: 4 discrete cases, each requiring different action. Additionally, a few individuals have a rare allergy to if-else statements that exceed 3 logical tests.
For more information on JavaScript's switch statement and its performance vs. if-else, please take a look at this post: Javascript switch vs. if...else if...else, which links to this more informative page http://archive.oreilly.com/pub/a/server-administration/excerpts/even-faster-websites/writing-efficient-javascript.html
You are slicing it wrong.
Use this code:
//_________________________________________________BEGIN notes
// Step 1. Get length of array
// Step 2. Find mid point
// Step 3. Compare if mid point is lower or higher than searched number
// Step 4. lop off unneeded side
// Step 5. go to step 1
//_________________________________________________END notes
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 44, 55];
function getMidPoint(arr, searchNumb) {
var length = arr.length;
var midPoint = Math.floor(length / 2);
var newArr = arr;
console.log(arr);
console.log("array midpoint value: " + arr[midPoint]);
if (arr[midPoint] > searchNumb) {
var newArr = arr.slice(0, midPoint);
return getMidPoint(newArr, searchNumb);
} else if (arr[midPoint] < searchNumb) {
var newArr = arr.slice(midPoint + 1, arr.length);
return getMidPoint(newArr, searchNumb);
} else {
return midPoint;
}
}
Also, if the search element is not in array, this will go on infinitely. Add a base case for that too.
I think that this line:
var newArr = arr.slice(0, arr[midPoint]);
should probably be:
var newArr = arr.slice(0, midPoint);
But I don't know if that's the only issue with your code. (It's not clear to me what the code is supposed to actually do. Right now "getMidPoint" appears to returns a smaller array containing the searched-for value.)
Probably You are already a master with Binary search. However I would like to indicate that is not necessary to create a sliding window for resolving a binary search.
function binarySearch(arr, value){
if(!arr.length) return -1;
let average = Math.floor(arr.length-1/2);
if (value === arr[average]) return average;
if (value > arr[average]) return binarySearch(arr.slice(average+1),value);
if (value < arr[average]) return binarySearch(arr.slice(0,average),value);
}
binarySearch([1,2,3,4,5],6) //-1
binarySearch([1,2,3,4,5],3) //2
Follow this steps to create the Binary search with recursion:
function binarySearch(arr, value){
1 ) implement a base case
if(!arr.length) return -1;
2 ) create a middle point
let average = Math.floor(arr.length-1/2);
3 ) if the middle point is equal to the searched valued, you found it! return the value
if (value === arr[average]) return average;
4) if the value is greater than the middle point run a new process with only the sub array starting from the middle + 1 till the end
if (value > arr[average]) return binarySearch(arr.slice(average+1),value);
5) if the value is lower than the middle point run a new process with only the sub array starting from 0 to the middle
if (value < arr[average]) return binarySearch(arr.slice(0,average),value);
}
I hope it helps!
Note: you can use a switch statement in order to not repeat if,if,if but I like it more this way, more readable.
For solving the question in recursion please find the answer and explanation below.
const BinarySearchRec = (arr, el) => {
// finding the middle index
const mid = Math.floor(arr.length / 2);
if (arr[mid] === el) {
// if the element is found then return the element.
return mid;
}
if (arr[mid] < el && mid < arr.length) {
/** here we are having the value returned from recursion as
the value can be -1 as well as a value which is in second half of the original array.**/
const retVal = BinarySearchRec(arr.slice(mid + 1, arr.length), el);
/** if value is greater than or equal to 0 then only add that value with mid
and also one as mid represents the index.
Since index starts from 0 we have to compensate it as we require the length here.**/
return retVal >= 0 ? mid + 1 + retVal : -1;
}
if (arr[mid] > el) {
// here we need not do any manipulation
return BinarySearchRec(arr.slice(0, mid), el);
}
return -1;
};
The above solutions which have been added and the one accepted fails in scenarios when the element to be found is in the second half.
There is solution with while loop which works correctly but since the question was to solve it recursively I have given a comprehensive recursive version.
There are 2 issues in your code :-
1) You are slicing it incorrectly
2) You have not put any base condition
This code should work hopefully :-
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 44, 55];
function getMidPoint(arr, searchNumb) {
var length = arr.length;
var midPoint = Math.floor(length / 2);
var newArr = arr;
console.log(arr);
console.log("array midpoint value: " + arr[midPoint]);
if (arr[midPoint] > searchNumb) {
var newArr = arr.slice(0, midPoint);
return getMidPoint(newArr, searchNumb);
} else if (arr[midPoint] < searchNumb) {
var newArr = arr.slice(midPoint+1, arr.length);
return getMidPoint(newArr, searchNumb);
} else {
return arr[midPoint];
}
}
This function would return undefined if element is not found in array.
This is fully rewritten code to achieve your goal (commented, linted).
This example doesn't have any checks for params.
Main error:
wrong slicing
Disadvantages of this approach:
recursion is slower and takes up more of the stack
slice() also there is no needed (because of the stack again)
/**
* Searches recursively number from the list
* #param {Array} list
* #param {number} item Search item
* #param {number} low Lower limit of search in the list
* #param {number} high Highest limit of search in the list
* #param {number} arrLength Length of the list
* #return {(number | null)} Number if the value is found or NULL otherwise
*/
const binarySearch = ( list, item, low, high, arrLength ) => {
while ( low <= high ) {
let mid = Math.floor((low + high) / 2);
let guess = list[mid];
if ( guess === item ) {
return mid;
} else if ( guess > item ) {
high = mid - 1;
list = list.slice( 0, mid );
return binarySearch( list, item, low, high );
} else {
low = mid + 1;
list = list.slice( low, arrLength );
return binarySearch( list, item, low, high );
}
}
return null;
};
/**
* Creates the array that contains numbers 1...N
* #param {number} n - number N
* #return {Array}
*/
const createArr = ( n ) => Array.from({length: n}, (v, k) => k + 1);
const myList = createArr( 100 );
const arrLength = myList.length;
let low = 0;
let high = arrLength - 1;
console.log( '3 ' + binarySearch( myList, 3, low, high, arrLength ) ); // 2
console.log( '-1 ' + binarySearch( myList, -1, low, high, arrLength ) ); // null
I think it's more elegant solution for binary search:
const binarySearch = ( list, item ) => {
let low = 0;
let high = list.length - 1;
while ( low <= high ) {
let mid = Math.floor((low + high) / 2);
let guess = list[mid];
if ( guess === item ) {
return mid;
} else if ( guess > item ) {
high = mid - 1;
} else {
low = mid + 1;
}
}
return null;
};
const myList = [1, 3, 5, 7, 9];
console.log( binarySearch( myList, 3 ) );
console.log( binarySearch( myList, -1 ) );
Here's my recursive binary search solution:
// arr = sorted array, val = search value
// left and right are the index pointers enclosing the search value
// e.g. binarySearch([1,5,7,9,14,17,24,29,33,38,49,52,61,62,70,80,90,95,104,107,109],70)
binarySearch = (arr,val,left=0,right=arr.length) => {
position = (left,right) => {
let pos = (left + right)/2
return Math.floor(pos)
}
let i = position(left,right)
if (arr[i] === val) {
return i
}
// Base Case: if left and midpoint index coincide then there are no more possible solutions
else if (i === left) {
return -1
}
// For this case we shift the left index pointer
else if (arr[i] < val) {
return binarySearch(arr,val,i,right)
}
// For this case we shift the right index pointer
else if (arr[i] > val) {
return binarySearch(arr,val,left,i)
}
}
Here is my approach for binary search recursively.
We don't slice the array because it is not needed if we can just pass down the indexes.
I think that will save some time.
Function will return index if the element is found and -1 if not.
l is standing for left, r is standing for right.
function binarySearch(arr, searchNumber) {
return _binarySearch(0, arr.length -1, arr, searchNumber);
function _binarySearch(l, r, arr, searchNumber) {
const mid = Math.floor((l + r) / 2);
const guess = arr[mid];
if (guess === searchNumber) { // base case
return mid;
} else if (l === r) { // end-case the element is not in the array
return -1;
} else if (guess < searchNumber) {
return _binarySearch(mid + 1, arr.length - 1, arr, searchNumber);
} else if (guess > searchNumber) {
return _binarySearch(l, mid - 1, arr, searchNumber);
}
}
}
const list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
console.log(binarySearch(list, 4));
Simple and Easy
let arr = [1,2,3,4,5];
function BinarySearch(arr, start, end, key) {
if(start > end) return -1;
let mid = Math.floor((start + end) / 2);
if(arr[mid] === key) return mid;
if(key > arr[mid]) {
return BinarySearch(arr, mid + 1, end, key);
} else if(key < arr[mid]) {
return BinarySearch(arr, start, mid -1, key);
}
}
BinarySearch([1,3,4,5], 0, arr.length - 1, 1); // it will return 0;
BinarySearch recursion Returning search element index.
Below code worked for me
function binerySearchRecursive(arr, num, start=0 end=arr.length-1){
let mid = Math.floor((start+end/2));
if(start> end){
return -1; // edge case if array has 1 element or 0
}
if(num === arr[mid])
return mid;
else if(num < arr[mid])
return binerySearchRecursive(arr, num, start, mid-1 );
else
return binerySearchRecursive(arr, num, mid+1 , end);
}
binerySearchRecursive([1,2,3,4,5], 5)
function binarySearch(arr, n) {
let mid = Math.floor(arr.length / 2);
// Base case
if (n === arr[mid]) {
return mid;
}
//Recursion
if (n > arr[mid]) {
return mid + binarySearch(arr.slice(mid, arr.length), n)
} else {
return binarySearch(arr.slice(0, mid), n)
} }
Simple solution to recursive binary search
For a recursive binary search you can try this :
function recursiveBinarySearch(lst, target, start=0, end=(lst.length-1)){
let midPoint = (Math.floor((start+end)/2));
if (start > end){
return false;
}
if (lst[midPoint] === target){
return true;
}
else{
if(lst[midPoint] < target){
return recursiveBinarySearch(lst, target, midPoint+1, end);
}
else{
return recursiveBinarySearch(lst, target, start, midPoint-1);
}
}
}
this too late but i hope this well be useful for some one :)
const items = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
let target = 30;
function binarySearch(L,R){
if(L == R){
return false;
}
let mid = Math.floor((L + R)/2);
if(mid == target){
return target;
}
if(mid > target){
binarySearch(L,mid);
}
if(mid < target){
binarySearch(mid+1,R);
}
}
binarySearch(1,items.length);
This is the most comprehensive version of binary recursive search for JavaScript. In my opinion, this is O(log n).
function binaryRecursion(arr, val) {
if (arr.length === 0) return -1
let middle = Math.floor(arr.length - 1 / 2)
if (arr[middle] === val) return middle;
if (val > arr[middle]) {
return binaryRecursion(arr.slice(middle + 1), val)
}
if (val < arr[middle]) {
return binaryRecursion(arr.slice(0, middle), val)
}
}
This returns the index of the element, not whether it exists or not.